Question

. In the figure AD||BC , X and Y are midpoints of AB and CD show that XY parallel to AD and BC and XY=½(AD+BC)

Answer 1

Given :- In the figure AD||BC , X and Y are midpoints of AB and CD .

To Show :-

• XY parallel to AD and BC .
• XY = ½(AD + BC) .

Solution :-

Construction :-

• Extand AY and BC such that , they meet at point P .

now, in ∆ADY and ∆PCY we have,

→ ∠AYD = ∠PYC { Vertically opposite angles .}

→ DY = CY { given. }

→ ∠ADY = ∠PCY { Alternate angles .}

so,

→ ∆ADY ≅ ∆PCY {By ASA congruence rule.}

then,

→ AD = PC { By CPCT. } ---------- Eqn.(1)

also,

→ AY = PY { By CPCT. } ---------- Eqn.(2)

now, in ∆ABP we have,

→ AX = XB { given. }

→ AY = YP { from Eqn.(2) }

therefore,

→ XY ll BP --------- Eqn.(3)

and,

→ XY = (1/2)BP { By mid point theorem. }

→ XY = (1/2)[BC + CP]

using Eqn.(1),

→ XY = (1/2)[AD + BC] [Proved.]

using Eqn.(3) now,

→ XY || BC

since,

→ AD || BC { given.}

hence,

→ XY || AD || BC [Proved.]

Learn more :-

in triangle ABC seg DE parallel side BC. If 2 area of triangle ADE = area of quadrilateral DBCE find AB : AD show that B...

https://brainly.in/question/15942930

2) In ∆ABC seg MN || side AC, seg MN divides ∆ABC into two parts of equal area. Determine the value of AM / AB

https://brainly.in/question/37634605